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$f$ is a continuous function on $[0,1]$.

Let $S(n)=\int_0^1x^nnf(x)dx$.

What is the limit of $S(n)$ as $n$ goes to infinity?

I have tried using product rule, but I have not made any progress.

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tony
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    Please fix your post by using MathJax to clearly write the quantities you have. – Ninad Munshi Dec 06 '19 at 09:56
  • What is "(x^n)nf"? – Elias Dec 06 '19 at 09:59
  • @AnonymousIGuess (x^n)nf(x) – tony Dec 06 '19 at 10:00
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    @lioness99a I know you mean well but looking back at OP's post history, it seems that they have very rarely edited their own posts with MathJax, having other people do it for them. – Ninad Munshi Dec 06 '19 at 10:03
  • @tony Okay. Try to use the fact that $f$ is continuous over the compact set $[0,1]$, which means that it is bounded there by the extreme value theorem. – Elias Dec 06 '19 at 10:05
  • @NinadMunshi Honestly, it takes such a short length of time to make the change that I just do it. I would rather spend 30 seconds fixing posts than looking at hundreds of unformatted questions – lioness99a Dec 06 '19 at 10:05
  • Another one: https://math.stackexchange.com/q/678024. – Martin R Dec 06 '19 at 10:44

1 Answers1

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Use the substitution $u=x^{n+1}$

$$S_n = \int_0^1 nx^n f(x)dx=\frac{n}{n+1}\int_0^1 f\left(u^{\frac{1}{n+1}}\right)du$$

Since $f$ is continuous on a compact set, it must be bounded, thus we can apply the Dominated Convergence Theorem:

$$\lim_{n\to\infty} S_n = \int_0^1 f\left(u^0\right)du = f(1)\int_0^1 du = f(1)$$

Ninad Munshi
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